The challenge

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

Constraints:

  • The number of nodes in the tree is in the range [0, 10<sup>5</sup>].
  • -1000 <= Node.val <= 1000

The definition for a binary tree node

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 public class TreeNode {
     int val;
     TreeNode left;
     TreeNode right;
     TreeNode() {}
     TreeNode(int val) { this.val = val; }
     TreeNode(int val, TreeNode left, TreeNode right) {
         this.val = val;
         this.left = left;
         this.right = right;
     }
 }

The solution in Java code

First we need to make sure that we catch our 2 edge cases:

  • root is null
  • root has no child nodes

Then we can check if left is null, it means that we can run the depth again, but with right, adding 1 level.

The same then applies to the right node.

Once this recursion is out, we make sure to use the built-in Math.min() to do the same logic if not caught above:

  • minimum value will be the left node under recursion
  • compared to the right node under recursion

Making sure to increment our final output by 1, as we have not yet counted in the additional level yet.

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class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) return 0;
        if (root.left == null && root.right == null) return 1;
        if (root.left == null) {
            return this.minDepth(root.right) + 1;
        }
        if (root.right == null) {
            return this.minDepth(root.left) + 1;
        }
        return Math.min(this.minDepth(root.left), this.minDepth(root.right)) +1;
    }
}